Question

The 5 terms of an AP are a1, a2, a3, and a5. Given:a1+a3+a5=-12 and a1. a2. a3=8.
Find the first term and the common difference.

Answer 1

Answer:

Let the series of in AP a1, a2, a3, a4, a5 be:

a-2d , a-d , a , a+d, a+2d

Given,

a1+a3+a5 = -12

a-2d+a+a+2d = -12

3a = -12

a= -4

Also given,

(a1)(a2)(a3) = 8

(a-2d)(a-d)(a) = 8

Substituting the value of a and solving for d we get 2 possible values for d

They are d=-3 (or) d=-5

Substituting and finding the values using d=-3 and a=-4

a2 = a-d = -4+3 = -1

a4 = a+d = -4–3 = -7

a5 = a+3d = -4-9 = -13

Sum = -1–7–13 = -21

Substituting and finding the values using d=-5 and a=-4

a2 = a-d = -4+5 = 1

a4 = a+d = -4–3 = -9

a5 = a+3d = -4–15 = -19

Sum = 1–9–19 = -27

boone chance :)

Answer 2

Answer:

a = 2 and d= - 3

Step-by-step explanation:

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